codeforces1016 D. Vasya And The Matrix（思维+神奇构造）

D. Vasya And The Matrix

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!

Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.

Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.

Input

The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.

The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.

The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.

Output

If there is no matrix satisfying the given constraints in the first line, output "NO".

Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.

If there are several suitable matrices, it is allowed to print any of them.

Examples

input

Copy

2 3
2 9
5 3 13

output

Copy

YES
3 4 5
6 7 8

input

Copy

3 3
1 7 6
2 15 12

output

Copy

NO
题意：题意：给出n，m表示有一个n*m的矩阵，然后第一行给出n个数，每个数ai表示第i行所有的数的亦或和，第二行给出m个数，每个数bi表示第i列所有数的亦或和。问，是否可以构造出这样的一个矩阵，如果可以，输出“YES”并且输出这个矩阵，否则，输出“NO”.
分析：这题是迷之构造法，我们只要考虑矩阵最后一列与最后一行便可以了，其他的为0；

#include<stdio.h>
const int maxn = 101;
int a[maxn],b[maxn],G[maxn][maxn];
int main( )
{
    int n,m,cnt1=0,cnt2=0;
    scanf("%d%d",&n,&m);
    for(int i=1 ; i<=n ; i++)
    {
        scanf("%d",&a[i]);
    }

    for(int i=1 ; i<=m ; i++)
    {
        scanf("%d",&b[i]);
    }

    cnt1=a[1],cnt2=b[1];
    for(int i=2 ; i<n ; i++)
    cnt1^=a[i];
    cnt1^=b[m];
    for(int i=2 ; i<m ; i++)
    cnt2^=b[i];
    cnt2^=a[n];
    if(cnt1!=cnt2)
    {
        puts("NO");
        return 0;
    }
    puts("YES");
    for(int i=1 ; i<n ; i++)
    for(int j=1 ; j<m ; j++)
    G[i][j]=0;
    for(int i=1 ; i<n ; i++)
    {
        G[i][m]=a[i];
    }
    for(int i=1 ; i<m ; i++)
    {
        G[n][i]=b[i];
    }
    G[n][m]=cnt1;
    for(int i=1 ; i<=n  ;i++)
    {
        for(int j=1 ; j<=m ; j++)
        printf("%d ",G[i][j]);
        puts(" ");
    }
    return 0;
}